Problem: A particle with velocity $v(t)=\sin(t)$, where $t$ is time in seconds, moves in a straight line. How far does the particle move from $t=0$ to $t=\dfrac{\pi}{2}$ seconds?
The definite integral $ \int_a^b |v(t)| \,dt$ gives the total distance traveled by a particle over the interval $[a,b]$. In this case, $ \int_0^{{\frac{\pi}{2}}} |\sin(t)| \,dt$ represents the total distance traveled by the particle from $t=0$ to $t=\dfrac{\pi}{2}$ seconds. Since $\sin(t)$ is always nonnegative on the interval $\Big[0,\dfrac{\pi}{2}\Big]$, we can rewrite the integral: $ \int_0^{{\frac{\pi}{2}}} \sin(t) \,dt$ We can now evaluate the integral: $\begin{aligned} \int_0^{{\frac{\pi}{2}}} \sin(t) \,dt&=\Big[-\cos(t)\Big]_0^{\frac{\pi}{2}}\\ \\ \\ &=-\cos\left(\dfrac{\pi}{2}\right)-(-\cos(0))\\ \\ &=0-(-1)\\ \\ &=1 \end{aligned}$ The answer: $1$ unit